Chapter 6 Lines and Angles Ex 6.1 Solutions
Question - 1 : - In the given figure, lines AB and CD intersect at O.If 
and 
find ∠BOE and reflex ∠COE.
Answer - 1 : -
Question - 2 : - In the given figure, linesXY and MN intersect at O. If ∠POY= 
and a:b =2 : 3, find c.
Answer - 2 : -
Let the common ratiobetween a and b be x.
∴ a =2x, and b =3x
XY is astraight line, rays OM and OP stand on it.
∴ ∠XOM + ∠MOP + ∠POY = 180º
b + a + ∠POY = 180º
3x + 2x +90º = 180º
5x = 90º
x = 18º
a = 2x = 2 × 18 =36º
b = 3x= 3 ×18 = 54º
MN is astraight line. Ray OX stands on it.
∴ b + c = 180º (Linear Pair)
54º + c =180º
c = 180º − 54º = 126º
∴ c =126º
Question - 3 : - In the given figure, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT.
Answer - 3 : -
In the given figure, ST isa straight line and ray QP stands on it.
∴ ∠PQS + ∠PQR =180º (Linear Pair)
∠PQR = 180º − ∠PQS (1)
∠PRT + ∠PRQ = 180º (Linear Pair)
∠PRQ = 180º − ∠PRT (2)
It isgiven that ∠PQR = ∠PRQ.
Equatingequations (1) and (2), we obtain
180º − ∠PQS = 180
− ∠PRT
∠PQS = ∠PRT
Question - 4 : - In the given figure, if 
then prove that AOB is a line.
Answer - 4 : -
It can be observed that,
x + y + z+ w =360º (Complete angle)
It is given that,
x + y = z + w
∴ x + y + x + y =360º
2(x+ y) = 360º
x + y = 180º
Since x and y forma linear pair, AOB is a line.
Question - 5 : - In the given figure, POQis a line. Ray OR is perpendicular to line PQ. OS is another ray lying betweenrays OP and OR. Prove that
Answer - 5 : -
It is given that OR ⊥ PQ
∴ ∠POR = 90º
⇒ ∠POS + ∠SOR = 90º
∠ROS = 90º − ∠POS … (1)
∠QOR = 90º (As OR ⊥ PQ)
∠QOS − ∠ROS = 90º
∠ROS = ∠QOS − 90º … (2)
On addingequations (1) and (2), we obtain
2 ∠ROS = ∠QOS − ∠POS
∠ROS = 
(∠QOS− ∠POS)
Question - 6 : - It is given that ∠XYZ= 
and XY is produced topoint P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Answer - 6 : - 
It is given that line YQbisects ∠PYZ.
Hence, ∠QYP = ∠ZYQ
It can beobserved that PX is a line. Rays YQ and YZ stand on it.
∴ ∠XYZ + ∠ZYQ + ∠QYP = 180º
⇒ 64º + 2∠QYP = 180º
⇒ 2∠QYP = 180º − 64º = 116º
⇒ ∠QYP = 58º
Also, ∠ZYQ = ∠QYP = 58º
Reflex ∠QYP =360º − 58º = 302º
∠XYQ = ∠XYZ + ∠ZYQ
=64º + 58º = 122ºIn the given figure, findthe values of x and y and then show that AB || CD.
Answer - 7 : -
It can be observed that,
50º + x = 180º (Linear pair)
x = 130º … (1)
Also, y = 130º (Vertically opposite angles)
As x and y arealternate interior angles for lines AB and CD and also measures of these anglesare equal to each other, therefore, line AB || CD.
Question - 8 : - In the given figure, if AB|| CD, CD || EF and y: z = 3: 7, find x.
Answer - 8 : -
It is given that AB || CDand CD || EF
∴ AB || CD || EF (Lines parallel to the sameline are parallel to each other)
It can beobserved that
x = z (Alternateinterior angles) … (1)
It isgiven that y: z = 3: 7
Let thecommon ratio between y and z be a.
∴ y =3a and z =7a
Also, x + y =180º (Co-interior angles on the same side of the transversal)
z + y = 180º [Usingequation (1)]
7a + 3a =180º
10a = 180º
a = 18º
∴ x =7a = 7 × 18º = 126º
Question - 9 : - In the given figure, If AB|| CD, EF ⊥ CD and ∠GED = 126º, find ∠AGE, ∠GEF and ∠FGE.
Answer - 9 : -
It is given that,
AB || CD
EF ⊥ CD
∠GED = 126º
⇒ ∠GEF + ∠FED =126º
⇒ ∠GEF + 90º = 126º
⇒ ∠GEF = 36º
∠AGE and ∠GED are alternate interior angles.
⇒ ∠AGE = ∠GED =126º
However, ∠AGE + ∠FGE = 180º (Linear pair)
⇒ 126º + ∠FGE = 180º
⇒ ∠FGE = 180º − 126º = 54º
∴ ∠AGE = 126º, ∠GEF =36º, ∠FGE = 54º
Question - 10 : - In the given figure, if PQ|| ST, ∠PQR =110º and ∠RST =130º, find ∠QRS.
[Hint: Draw a line parallel to STthrough point R.]
Answer - 10 : -
Let us draw a line XYparallel to ST and passing through point R.
∠PQR + ∠QRX = 180º (Co-interior angles on the same side of transversal QR)
⇒ 110º + ∠QRX = 180º
⇒ ∠QRX = 70º
Also,
∠RST + ∠SRY = 180º (Co-interior angles on the same side of transversal SR)
130º + ∠SRY = 180º
∠SRY = 50º
XY is astraight line. RQ and RS stand on it.
∴ ∠QRX + ∠QRS + ∠SRY = 180º
70º + ∠QRS + 50º = 180º
∠QRS = 180º − 120º = 60º